1
\$\begingroup\$

The formula I've seen is to multiply the signal by its complex conjugate, then integrate across the time period interval, and then divide by the period. But when I do that I don't get A^2/2

enter image description here

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Okay. So why don't you get \$\displaystyle \frac{A^2}{2}?\$ It would be helpful to know what you did so that we can tell you where you were incorrect. \$\endgroup\$ – KingDuken Mar 25 at 20:24
  • \$\begingroup\$ hi i posted my work and where im stuck \$\endgroup\$ – David Mar 25 at 20:33
  • \$\begingroup\$ im confused...shouldn't those terms be 0 if the final answer is to be A^2/2? \$\endgroup\$ – David Mar 25 at 20:40
  • \$\begingroup\$ isn't that the formula for power though? \$\endgroup\$ – David Mar 25 at 20:46
  • \$\begingroup\$ Sorry, misread the question AGAIN. I deleted my wrong comments. \$\endgroup\$ – DKNguyen Mar 25 at 20:50
1
\$\begingroup\$

The first image you posted appears to use Parseval's Relation after expressing the \$S_i(t)\$ as an exponential Fourier series.

Your method (the second image, I think) will produce the same answer, as long as you're careful about your limits of integration. Let's assume that you've defined \$T_0 = \frac{2\pi}{\omega_c+\omega_m}\$. Then your underlined terms can be recombined (using Euler "in reverse") to produce a cosine with frequency \$2(\omega_c+\omega_m)\$. This cosine will complete two full oscillations in \$T_0\$ seconds, so its integral is zero.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.